CBSE Class 11-science Answered

Consider two bodies A and B of masses m₁ and m₂ respectively, moving along the same straight line, in the same direction. Let be their respective velocities, such that

The two bodies will collide after some time.

During collision, the bodies will be deformed at the region of contact. So, a part of the kinetic energy will be converted into potential energy. The bodies will regain their original shape due to elasticity. The potential energy will be reconverted into kinetic energy. The bodies will separate and continue to move along the same straight line, in the same direction, but with different velocities.

Applying the law of conservation of momentum,

Total momentum before collision = total momentum after collision

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f (in magnitude)

or m₁ (v_1i - v_1f) = m₂ (v_2f - v_2i) --------------- (i)

Since the collision is elastic, the kinetic energy will be conserved.

Kinetic energy before collision = Kinetic energy after collision

Dividing (ii) by (i), we get

v_1i + v_1f = v_2f + v_2i

or v_1i - v_2i = v_2f - v_1f ------------ (iii)

(v_1i - v_2i) is the magnitude of the relative velocity of A w.r.t B.

(v_2f - v_1f) is the magnitude of the relative velocity of B w.r.t A.

It may be noted that the direction of relative velocity is reversed after the collision.

From equation (iii)

v_2f = v_1i - v_2i + v_1f

From equation (i)

m₁ (v_1i - v_1f) = m₂ (v_1i - v_2i + v_1f - v_2i)

or m₁v_1i - m₁v_1f = m₂v_1i - 2m₂v_2i + m₂v_1f

or -m₁v_1f - m₂v_1f = -m₁v_1i + m₂v_1i - 2m₂v_2i

or (m₁ + m₂) v_1f = (m₁ - m₂) v_1i + 2m₂v_2i

Again, from equation (iii),

v_1f = v_2f - v_1i + v_2i

Substituting this value in equation (i) and simplifying, we get

Hope this clarifies your doubt.

Regards

Team

Topperlearning